Completing a basis for $\mathbb{R}^4$ Find a basis for $\mathbb{R}^4$ containing the vectors  $(0, 1, 2, 3)$ and $(0, 1, 0, 1)$. 
Having trouble with this; I know that there are 4 vectors in the basis, and that any coordinate in $\mathbb{R}^4$ can be represented by a linear combination of them, but I'm not sure how I should go about finding them. 
 A: Hint.
One can prove that in order to complete an independent family of vectors to a basis, you can pick-up vectors from another basis.
Using this result, you can use vectors from the canonical basis to complement your initial family and look one by one to see if you still get an independant by adding one and then a second vector to get at the end four vectors.
Application of the method
As the first coordinate of your two initial vectors are equal to $0$, the vector $e_1=(1,0,0,0)$ from the canonical basis can be added to those to keep an independent family.
You then need for the last step to pick-up a vector from $\{e_2,e_3,e_4\}$ to complement the three previous vectors to get an independent family of $4$ vectors and therefore a basis. You can verify that $e_3=(0,0,1,0)$ works.
A: You should definitely have $(1,0,0,0)$ because your two current vectors cannot generate anything non-zero for the first element.
Then you would find a normal vector to $(1,2,3)$ and $(1,0,1)$ which is the cross product of the two: $(2,2,-2)$ and extending it to the fourth dimension gives $(0,2,2,-2)$
A: Put your two vectors as two rows of a $4\times4$ matrix. Randomly make two more rows. It's highly like (though not certain) that your four rows form a basis. Compute the determinant to check, and if necessary, try again. (The vectors form a basis if the determinant is nonzero.) Use a lot of $0$s to make the determinant easy to calculate.
